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3 years ago

5

out of 35 librarians surveyed 12 librarians chose fantasy as their favorite 9 chose nonfiction and 14 chose fiction qhat percent

chose fiction

1 answer:

Setler [38]3 years ago

7 0

14/35 = 0.4
0.4 X 100= 40%

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Evan and Eileen Carter are husband and wife and file a joint return for 2019. Both are under 65 years of age. They provide more

S_A_V [24]

Answer:

The correct answer is b) four.

Step-by-step explanation:

Evan and Eileen Carter must submit a separate statement as dictated by the requirements for the claim.

In the case of the daughter, she no longer meets the requirement of a qualified child because she is over 24 years old; but qualifies as an eligible relative, and also, the income of her scholarship is not enough, so she needs the help of her parents to cover her expenses.

The couple also provides support in all expenses of Evan's great-grandmother; although Belinda lives in an asylum, the Carter can claim the exemption since, for qualified relatives, they do not ask as a mandatory requirement that they live in the same place.

And finally, Peggy cannot request an exemption since she is not a qualified relative, so she does not fall within the exceptions to claim, even if she lives in Evan and Eileen Carter´s house.

Therefore, the Carter can claim four dependency exemptions, one for each person that are Evan, Eileen, Pamela, and Belinda.

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<em>I hope this information can help you.</em>

7 0

4 years ago

Bess [88]

It's forming a circle so we know that together it's going to add up to 360. #1 is 66 because angle A trough E is straight and we have 90 on angle ACF and it's vertical to angle FCG. #2 is 125 because angle ACB is 24 because it's vertical to angle GCE. #3 is 114 because 90 and 24 is 114. #4 is 156 because 66 and 90 equal 156.
1. FCG = 66
2. BCD = 125
3. FCB = 114
4. ACG = 156

4 0

4 years ago

Perform the indicated operation. Be sure the answer is reduced.

avanturin [10]

<h3>Given Equation:-</h3>

$\boxed{ \rm \frac{4x^{2}y^{3}z}{9} \times \frac{45y}{8 {x}^{5} {z}^{5} }}$

<h3>Step by step expansion:</h3>

$\dashrightarrow \sf\dfrac{4x^{2}y^{3}z}{9} \times \dfrac{45y}{8 {x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{ \cancel4x^{2}y^{3}z}{9} \times \dfrac{45y}{ \cancel8 {x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{9} \times \dfrac{45y}{2{x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{ \cancel9} \times \dfrac{ \cancel{45}y}{2{x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{1} \times \dfrac{5y}{2{x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{0}y^{3}z}{1} \times \dfrac{5y}{2{x}^{5 - 2} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{y^{3}z}{1} \times \dfrac{5y}{2{x}^{3} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{y^{3}z {}^{0} }{1} \times \dfrac{5y}{2{x}^{3} {z}^{3 - 1} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{y^{3}}{1} \times \dfrac{5y}{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \sf \dfrac{5y \times {y}^{3} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \sf \dfrac{5y {}^{0} \times {y}^{3 + 1} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \sf \dfrac{5 \times {y}^{4} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \bf \dfrac{5 {y}^{4} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\therefore \underline{ \textbf{ \textsf{option \red c \: is \: correct}}}$

8 0

2 years ago

In a lottery the top cash prize and $663 million going to three lucky winners players pick five different numbers from 1 to 58 a

user100 [1]

Answer:

43

Step-by-step explanation:

5 0

4 years ago

If 12 +22+ 32 +

kakasveta [241]

Answer:

m2

Step-by-step explanation:

the 2nd equation is double the first one and the first one is 1

8 0

2 years ago